Hello all. Im bumping this cause i am looking into building something similar as well. Im quite new to the world of electronics and I want to make a dual output power supply going from +12v to 0 to -12v, to drive a DIY modular synth that uses op amps and so on. Anything that i came across uses step up transformers that go up to 240v that i don't need and honestly scare me a little. I got my hands on a couple of PSUs that output 13.5VDC and i want to invert this into +12v and -12v (hence the need for the inverter).

 

Rather than somehow trying to invert the power supply output voltage polarity, simply connect the supply output inverted? one supply with the negative output tied to "common, and the second supply with the positive output tied to that same common, and the negative side available?? That is the standard scheme to achieve a dual polarity DC supply.
Certainly somebody can produce that drawing.

And if the supplies do not have the output isolated from the input, I would not even consider using them.

 

Rather than somehow trying to invert the power supply output voltage polarity, simply connect the supply output inverted? one supply with the negative output tied to "common, and the second supply with the positive output tied to that same common, and the negative side available?? That is the standard scheme to achieve a dual polarity DC supply.
Certainly somebody can produce that drawing.

And if the supplies do not have the output isolated from the input, I would not even consider using them.

Thanks for the reply @MisterBill2.

Thankfully those wall adapters do isolate the input from the output and I am aware that connecting them in "series" will give me the result i want but im trying to find a solution only utilising one of them because i feel like plugging in 2 wall adapters for one synth is kind of impractical.

Im following the designs of a youtuber (Mortiz Klein), in a video he explained how to make a dual output PSU basically taking the AC signal from an AC output wall adapter rectifying that in both ways passing that through big capacitors and then voltage regulators giving him a +12 ,0 and -12 supply. Schematics included below.
(from my understanding his wall adapter just steps down the voltage from 240v to 15 or something like that but its still AC)

Since i dont have an AC wall adapter im trying to somehow convert the 13.5DC that i get from my wall adapter to AC in order to create the same circuit as Mortiz.

Im open to ideas on how i could i achieve this even if this strays from the original plan.

 

For a supply that is producing only 1 A, that is a *lot* of filter capacitance. Peak-to-peak ripple at the input to the regulators will be around 1.2 V. Is there a reason it has to be so low before regulation?

ak

https://forum.allaboutcircuits.com/threads/12v-dc-to-ac-converter-circuit-design-2.203857/

Apologies for the confusion in regards to the thread. @AnalogKid

About your last reply here, I can only guess because Im not really educated on this and Im just starting out on my DIY synth and electronics journey, that the peak to peak difference is so low for the voltage regulators to work more efficiently and produce less heat having less power loss in the end.

What would you suggest i use and what should i look for in order to calculate the capacitance on my own?
Also one last note, I will not be using just 2 diodes but a full bridge rectifier if that will make any difference.

 

Converting the DC to AC and then back to DC involves a bunch of unneeded circuitry and can add switching noise to the system.
Just put the two 13.5Vdc adapters in a small electronics box, along with two linear LDO (Low Drop Out) regulators to give plus and minus 12Vdc.
Run an extension cord into the box for the adapter power, so you only have one wall plug.

Below is the LTspice sim of an example circuit using a 12V LDO.

 

that the peak to peak difference is so low for the voltage regulators to work more efficiently and produce less heat having less power loss in the end.

<major edit from the first posting>

Actually, the exact opposite is true. The positive peaks are set by the transformer secondary peak voltage, so they are quasi-constant as the load current increases of decreases. But the negative peaks are caused directly by the load current; the higher the load current, the lower the negative peak voltage value. With the upper peaks a near constant, as the lower peals more lower, the average value of the ripple waveform decreases.

As a *very* crude approximation, let's say that at some load, the peak DC voltage across the filter cap is 20 V, and the ripple voltage dips down to 18 V. In this case the average voltage seen by the regulator is 19 V, and this determines the heat dissipated by the regulator. Now, if you remove 50% of the filter capacitance, the peak voltage remains at 20 V but the ripple voltage extends down to 16 V (for the same output current). Now the average voltage into the IC is 18 V, not 19 V. This reduces the IC power dissipation.

Of course, you have to make sure the ripple voltage negative peaks do not extend below the regulator's minimum input voltage spec.

ak

 

Thank you all for the advice. I just realised with a 13.5VDC wall adapter i still cant use the L7812 to output a steady 12v as it needs at least 14v. In mortiz's video he uses a 15VAC wall adapter so the ripple voltage makes sense to be relatively low as the L7812 needs the necessary headroom to operate, so it cant dip below 1v.

 

Welcome to AAC.

Something like this, which uses a TI buck-boost chip, is a good solution.

 

Thank you all for the advice. I just realised with a 13.5VDC wall adapter i still cant use the L7812 to output a steady 12v as it needs at least 14v. In mortiz's video he uses a 15VAC wall adapter so the ripple voltage makes sense to be relatively low as the L7812 needs the necessary headroom to operate, so it cant dip below 1v.

That's why I showed an LDO regulator which only needs about 1V drop to operate.

 

Im stuck at a predicament where i have a 13.5VDC wall adapter where firstly i need to first increase it by a few V (to at least give the IC the minimum voltage needed) and then invert it to Ac to finally feed it to the L7812.

The other option that I have is using another 3.3VAC wall adapter that i have and then I need to multiply the voltage to at least 14VAC and then filter it and feed it to the L7812.

The second option i think is off the table due to the very little current supply that il have.

I think by best option is @crutschow solution.